Fairly simple if you convert to modulus-argument form.

Hi Daniel123
this is the proof for natural n:
√3+i=2(√3/2  + (1/2)i)=2(cos 30° +i sin(30° ))  r=2 ,  theta=30° then by demoivers theorem

(√3+i)^n=2^n(cos n(30°) +i sin( n (30° ))  and similarly

√3-i=2(√3/2  - (1/2)i)=2(cos 330° +i sin(330° ))  r=2 ,  theta=30° then by demoivers theorem

(√3-i)^n=2^n(cos n(330°) +i sin( n (330° ))   therefore

(√3+i)^n+(√3-i)^n=2^n(cos n(30°) +i sin( n (30° ))+ 2^n(cos n(330°)+i sin( n (330° )) but
sin( n (30°))=- sin( n (330° )) therefore the imaginary part cancells and the result is in R or
(√3+i)^n+(√3-i)^n=2^n(cos n(30°)+2^n(cos n(330°) but cos n(30°)=(cos n(330°) i.e
(√3+i)^n+(√3-i)^n=2(2^n(cos n(30°))=2^(n+1)   (cos n(30°) ∈R
Best Regards
Riad Zaidan